## Numerical solution of integral equation with parametric variable

### Ann (view profile)

on 7 Apr 2011
Latest activity Edited by Pooyan

### Pooyan (view profile)

on 9 Jun 2014
Accepted Answer by Jarrod Rivituso

### Jarrod Rivituso (view profile)

Hello!

Please, how can I solve integral equation in Matlab

L = integral (f(b,t)) dt for third variable (parametric variable b)

Limits of integral are t0, t1.

bym

### bym (view profile)

on 7 Apr 2011

numerically or symbolically?

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### Jarrod Rivituso (view profile)

Answer by Jarrod Rivituso

### Jarrod Rivituso (view profile)

on 7 Apr 2011

OK, I'm gonna assume you want to do it numerically. Check out this (warning, it gets a little crazy with the function handles):

```function solveIntegralForB
```
```bfound = fsolve(@func2minimize,2)
```
```      function output = func2minimize(b)
t0 = 0;
t1 = 3;
L = 50;
output = (L - quad(@myFunc,t0,t1))^2;```
```          function f = myFunc(t)
f = exp(b*t);
end```
`      end`
```end
```

Essentially, what it does is use the quad function to perform an integration for some value of b. Additionally, it uses the fsolve function to then minimize the "func2minimize" function, which performs the integral for some value of b and checks it against my desired solution.

Here, I've assumed a simple function (exp(b*t)) as f, but you could see how it could be changed. Of course, this is an iterative solution, so there's no guarantee of it finding a solution for all functions f.

Hope this helps!

Pooyan

### Pooyan (view profile)

on 9 Jun 2014

How if L is a known function of t and b is an unknown function of t? Can anyone help me on that? and again I want to solve the equation for b(t). Let's assume L(t)= sin(t) and t=0:0.1:10;

Thanks.

### Matt Tearle (view profile)

Answer by Matt Tearle

### Matt Tearle (view profile)

on 8 Apr 2011

A variation on Jarrod's approach, using function handles (because everyone loves function handles):

```myFunc = @(t,b) exp(t*b); % or whatever
t0 = 0;
t1 = 3;
L = 50;
f = @(b) quad(@(t) myFunc(t,b),t0,t1);
bsolve = fzero(f,2);
```

Or fsolve instead of fzero if you have Optimization Toolbox.

### Walter Roberson (view profile)

Answer by Walter Roberson

### Walter Roberson (view profile)

on 11 Apr 2011

If you have the symbolic toolbox,

```syms x b
solve(int(f(x,b),x,t0,t1)-L,b)
```

In theory if it can be solved symbolically it will do so, and if not then MuPad should switch to numeric integrations, I think.